First slide

Introduction to mathematical reasoning

Question

Negation of $q \lor ~ (p \land r)$ is

Easy

A

$~ q \lor ~ (p \lor r)$
B

$~ q \land (p \land r)$
C

$~ q \land (p \land r)$ .
D

$~ q \lor (p \land r)$ .

Solution

$~ [q \lor ~ (p \land r)] \equiv\sim q \land (p \land r)$

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